Find original Array from given Array where each element is sum of prefix and postfix sum
Last Updated :
06 Nov, 2023
Given an array arr[] of length N, where arr is derived from an array nums[] which is lost. Array arr[] is derived as:
arr[i] = (nums[0] + nums[1] + … + nums[i]) + (nums[i] + nums[i+1] + … + nums[N-1]).
The task is to find nums[] array of length N.
Examples:
Input: N = 4, arr[] = {9, 10, 11, 10}
Output: {1, 2, 3, 2}
Explanation: If nums[] = {1, 2, 3, 2}, then according to above definition
arr[0] = (nums[0]) + (nums[0] + nums[1] + nums[2] + nums[3]) = 1 + 1 + 2 + 3 + 2 = 9
arr[1] = (nums[0] + nums[1]) + (nums[1] + nums[2] + nums[3]) = 1 + 2 + 2 + 3 + 2 = 10
arr[2] = (nums[0] + nums[1] + nums[2]) + (nums[2] + nums[3]) = 1 + 2 + 3 + 3 + 2 = 11
arr[3] = (nums[0] + nums[1] + nums[2] + nums[3]) + (nums[3]) = 1 + 2 + 3 + 2 + 2 = 10
Input: N = 2, arr[] = [25, 20]
Output: [10, 5]
Approach: Follow the below idea to solve the problem:
Suppose nums[] contains [a1, a2, a3, …, aN]
Then, sum = a1 + a2 + a3 + . . . + aN.
We are given
b1 = a1 + a1 + a2 + . . . + aN = a1 + sum …..(1)
Similarly,
b2 = a1 + a2 + a2 + . . . + aN = a2 + sum …..(2)
. . . (so on) and in last
b1 = a1 + a2 + a3 + . . . + aN + aN = aN + sum …..(N)
where [b1, b2, b3 , . . ., bN] are elements of arr[] and,
total = b1 + b2 + b3 + . . . + bN
Adding all equation (1) + (2) + (3) + …. + (N) we will get
b1 + b2 + b3 + . . . + bN = (a1 + sum) + (a2 + sum) + . . . + (aN + sum)
total = (a1 + a1 + a2 + . . . + aN) + (N * sum)
total = (sum) + (N * sum)
total = (N + 1) * sum
Now find the value of sum variable after that simply:
a1 = (b1 – sum), a2 = (b2 – sum), . . ., aN = (bN – sum)
Using the above idea follow the below steps to implement the code:
- First of all, try to store the sum of elements of arr[] in a variable let’s say total
- Using the formula (N + 1) * sum = total, we will get the value of variable sum which denotes the sum of elements present in the nums[] array.
- At last traverse N times to find nums[0] = arr[0] – sum, nums[1] = arr[1] – sum and so on.
- Return the array and print it.
Below is the implementation of the above approach:
C++
#include <iostream>
#include <vector>
using namespace std;
vector<int> findOrgArray(vector<int> arr, int N)
{
int total = 0;
for (int val : arr)
total += val;
int sum = (total / (N + 1));
vector<int> v;
for (int i = 0; i < N; i++) {
int val = arr[i] - sum;
v.push_back(val);
}
return v;
}
int main()
{
int N = 4;
vector<int> arr = { 9, 10, 11, 10 };
vector<int> v = findOrgArray(arr, N);
for (auto val : v)
cout << val << " ";
return 0;
}
|
Java
import java.util.*;
class GFG {
public static void main(String[] args)
{
int N = 4;
int[] arr = { 9, 10, 11, 10 };
List<Integer> nums = findOrgArray(arr, N);
for (int x : nums)
System.out.print(x + " ");
}
public static List<Integer> findOrgArray(int[] arr,
int N)
{
int total = 0;
for (int val : arr)
total += val;
int sum = (total / (N + 1));
List<Integer> nums = new ArrayList<>();
for (int i = 0; i < N; i++) {
int val = arr[i] - sum;
nums.add(val);
}
return nums;
}
}
|
Python3
def findOrgArray(arr, N) :
total = 0
for i in arr :
total+= i
sum = int(total / (N + 1));
v = []
for i in range (N) :
val = arr[i] - sum
v.append(val)
return v
if __name__ == "__main__" :
N = 4
arr = [ 9, 10, 11, 10 ]
v = findOrgArray(arr, N)
for val in v :
print(val,end=' ')
|
C#
using System;
using System.Collections.Generic;
public class GFG
{
public static List<int> findOrgArray(int[] arr,
int N)
{
int total = 0;
foreach (int val in arr)
total += val;
int sum = (total / (N + 1));
List<int> nums = new List<int>();
for (int i = 0; i < N; i++) {
int val = arr[i] - sum;
nums.Add(val);
}
return nums;
}
public static void Main(String []args)
{
int N = 4;
int[] arr = { 9, 10, 11, 10 };
List<int> nums = findOrgArray(arr, N);
for (int x = 0; x < nums.Count; x++)
Console.Write(nums[x] + " ");
}
}
|
Javascript
<script>
function findOrgArray(arr, N)
{
let total = 0;
for (let i = 0; i < N; i++)
total += arr[i];
let sum = (total / (N + 1));
let v= new Array(N);
for (let i = 0; i < N; i++) {
v[i] = arr[i] - sum;
}
return v;
}
let N = 4;
let arr = [ 9, 10, 11, 10 ];
let v = findOrgArray(arr, N);
for (let i = 0; i < N; i++)
document.write(v[i]+ " ");
</script>
|
Time Complexity: O(N)
Auxiliary Space: O(N), to further reduce it to O(1), store the value in the same given array arr[] rather than storing it in a new array.
Another Approach:
- Initialize a variable named “total” to 0.
- Traverse the input array “arr” using a range-based for loop.
a. For each element “val” in “arr”, add “val” to the “total” variable.
- Compute the sum of the original array “nums” using the formula: sum = total / (N + 1).
- Traverse the input array “arr” again using a for loop with index “i” from 0 to N-1.
a. For each element in “arr”, subtract “sum” from it and store the result back into “arr[i]”. This effectively undoes the modification made to “arr” and recovers the original array “nums”.
Below is the implementation of the above approach:
C++
#include <iostream>
#include <vector>
using namespace std;
void findOrgArray(vector<int>& arr, int N)
{
int total = 0;
for (int val : arr)
total += val;
int sum = (total / (N + 1));
for (int i = 0; i < N; i++) {
arr[i] = arr[i] - sum;
}
}
int main()
{
int N = 4;
vector<int> arr = { 9, 10, 11, 10 };
findOrgArray(arr, N);
for (auto val : arr)
cout << val << " ";
return 0;
}
|
Java
import java.util.*;
import java.io.*;
public class GFG {
static void findOrgArray(List<Integer> arr, int N) {
int total = 0;
for (int val : arr) {
total += val;
}
int sum = total / (N + 1);
for (int i = 0; i < N; i++) {
arr.set(i, arr.get(i) - sum);
}
}
public static void main(String[] args) {
int N = 4;
List<Integer> arr = new ArrayList<>();
arr.add(9);
arr.add(10);
arr.add(11);
arr.add(10);
findOrgArray(arr, N);
for (int val : arr) {
System.out.print(val + " ");
}
}
}
|
Python3
def find_org_array(arr, N):
total = 0
for val in arr:
total += val
sum = int(total / (N + 1))
for i in range(N):
arr[i] -= sum
return arr
arr=[9, 10, 11, 10]
n=4
arr=find_org_array(arr,n)
for val in arr:
print(val,end=" ")
|
C#
using System;
using System.Collections.Generic;
class Program
{
static void FindOrgArray(List<int> arr, int N)
{
int total = 0;
foreach (int val in arr)
total += val;
int sum = total / (N + 1);
for (int i = 0; i < N; i++)
{
arr[i] = arr[i] - sum;
}
}
static void Main()
{
int N = 4;
List<int> arr = new List<int> { 9, 10, 11, 10 };
FindOrgArray(arr, N);
foreach (var val in arr)
Console.Write(val + " ");
}
}
|
Javascript
function findOrgArray(arr, N) {
let total = 0;
for (let val of arr) {
total += val;
}
let sum = Math.floor(total / (N + 1));
for (let i = 0; i < N; i++) {
arr[i] = arr[i] - sum;
}
}
let N = 4;
let arr = [9, 10, 11, 10];
findOrgArray(arr, N);
for (let val of arr) {
console.log(val + " ");
}
|
Time Complexity: O(n)
Auxiliary Space: O(1)
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